Question

$\text{In an isosceles-trapezium, show that the opposite angles are supplementary.}$

Answer 1

Given : $ABCD$ is isosceles trapezium.

$AD=BC$

To Prove :

(i) $\angle A+\angle C={180}^{\circ }$

(ii)$\angle B+\angle D={180}^{\circ }$

Proof :

In isosceles trapezium $ABCD$, $AB\parallel CD$

then

$\angle D=\angle C$ [Since, base angles of isosceles trapezium are equal]

$\Rightarrow \angle A+\angle D={180}^{\circ }$ [Since, $AB\parallel CD$]

$\Rightarrow \angle A+\angle C={180}^{\circ }$ ---(1)

Now,

$\angle A+\angle C+\angle B+\angle D={360}^{\circ }$ [Since, sum of the angles of quadrilateral]

$\Rightarrow {180}^{\circ }+\angle B+\angle D={360}^{\circ }$

$\Rightarrow \angle B+\angle D={360}^{\circ }-{180}^{\circ }$

$\Rightarrow \angle B+\angle D={180}^{\circ }$ ---(2)

From (1) and (2),

Opposite angles of isosceles trapezium are supplementary.

Hence, proved.